How to show that $\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}=9/2$?

$\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}$

I need to prove that this limit equals to $\frac{9}{2}$. Can someone give me a step by step solution?

EDIT: I am sorry. The $x$ goes to $0$, not $1$.

If you allow Taylor expansions, recall that

$$\sin(x)=x-\frac16x^3+\mathcal O(x^5)$$

Thus,

$$\sin(3x)=3x-\color{red}{\frac92}x^3+\mathcal O(x^5)$$

Thus,

\begin{align}\frac{3x-\sin(3x)}{x^3}&=\frac{\frac92x^3+\mathcal O(x^5)}{x^3}\\&=\frac92+\mathcal O(x^2)\\&\to\frac92\end{align}

• Very nice solution, thanks a lot. – Thomas Feb 28 '17 at 23:50
• Taylor expansions make everything easy especially this limit as this limit has a question dedicated to it for proving without Taylor or L'hoptial. – A---B Mar 1 '17 at 0:02
• @A---B Indeed, but it's not to hard to derive the Taylor expansion either. Once you know the basic trig limits, deriving derivative of derivatives is not hard at all. – Simply Beautiful Art Mar 1 '17 at 0:09

Applying L'Hopital's rule three times $$\lim_{x \to 0}\frac{3x-\sin(3x)}{x^3}=\lim_{x \to 0}\frac{3-3\cos(3x)}{3x^2}=\lim_{x \to 0}\frac{9\sin(3x)}{6x}=\lim_{x \to 0}\frac{27\cos(3x)}{6}=\frac{9}{2}.$$

Using l'Hôpital's rule;$$\lim_{x\to0}\frac{3x-\sin(3x)}{x^3}=\lim_{x\to0}\frac{3-3\cos(3x)}{3x^2}=\lim_{x\to0}\frac{9\sin(3x)}{6x}=\lim_{x\to0}\frac{27\cos(3x)}{6}=\frac{9}{2}$$

• Yeah i tried to use l'Hôpital's rule but I did something wrong. Thanks for helping me out – Thomas Feb 28 '17 at 23:51

By elementary means:

From

$$\sin 3x=3\sin x-4\sin^3x$$

we draw

$$L=\lim_{x\to0}\frac{3x-3\sin x+4\sin^3x}{x^3}=\lim_{x\to0}\frac{3x-3\sin x}{x^3}+4.$$

But

$$\lim_{x\to0}\frac{x-\sin x}{x^3}=\lim_{3x\to0}\frac{3x-\sin3x}{27x^3}$$ so that

$$L=\frac L9+4.$$

• This doesn't prove that $L$ exists, does it? – user1551 Mar 1 '17 at 12:43
• @user1551: that's right, it only proves that if it exists, it has value $9/2$. – Yves Daoust Mar 1 '17 at 13:10

Hint: Apply the Hospital rule 3 times you obtain $\lim_{x\rightarrow 0}{{27\cos(3x)}\over 6}={9\over 2}$.

• No offense, but I think this is rather low quality. – Simply Beautiful Art Feb 28 '17 at 23:38
• Why so? Not agreeing or disagreeing, but could you elaborate just a bit? – Harnoor Lal Feb 28 '17 at 23:46

Use L'Hospital's rule:

$$\lim _{x\to0} \frac{3x-\sin3x}{x^3} = \lim_{x\to0} \frac{3-3\cos3x}{3x^2} = \lim _{x\to0} \frac{9\sin3x}{6x} = \lim_{x\to0} \frac{27\cos3x}{6} = \frac{27}{6} = \frac{9}{2}$$

• There is a symbol \to in $\LaTeX$ which renders as a right arrow: $\to$ and serves for limits instead of a 'minus'+'greater than' pair. – CiaPan Mar 1 '17 at 11:24
• I am rather new to LATEX so I am not quite acclimatized with it.Please bear with me. – Saradamani Mar 2 '17 at 4:01
• Don't worry, everyone was once a beginner :) – CiaPan Mar 2 '17 at 21:13

first :

theorem :

let $$f(0)=0 , f'(0)$$ have existed then :

$$\lim_{x\to 0} \frac{f(x)-\sin(f(x))}{x^3}= \frac{1}{6}(f'(0))^3$$

now :

$$\lim_{x\to 0} \frac{3x-\sin(3x)}{x^3}= \frac{1}{6}(3)^3=\frac{9}{2}$$